hmwk5sol2027 - Solutions to Homework 5, ISyE 2027 Problem 1...

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Problem 1 (a) Poisson (b) geometric (c) binomial (d) Bernoulli. Problem 2 (a) Notice that 1 = 4 X k =1 P ( X = k ) = 4 X k =1 ck = c (4)(5) 2 = 10 c so c = 1 / 10. (b) The expected value of X is just E [ X ] = 4 X k =1 kP ( X = k ) = 4 X k =1 (1 / 10) k 2 = 1 10 (4)(5)(9) 6 = 3 . (c) The range of Y is { 0 , 1 , 4 } , so we need to compute: P ( Y = 0) = P ( X = 2) = 1 / 5 P ( Y = 1) = P ( X = 1) + P ( X = 3) = 2 / 5 P ( Y = 4) = P ( X = 4) = 2 / 5 . Thus, the p.m.f. of Y is Pr { Y = k } = 1 / 5 for k = 0 2 / 5 for k = 1 2 / 5 for k = 4 0 otherwise. (d) The expected value of Y is E [ Y ] = (0) P ( Y = 0) + (1) P ( Y = 1) + (4) P ( Y = 4) = 2 / 5 + 4(2 / 5) = 2 . (e) Computing the conditional pmf of X , given X 2, means that we need to compute P ( X = 1 | X 2) and P ( X = 2 | X 2). Here P ( X = 1 | X 2) = P ( X = 1 ,X 2) P ( X 2) = P ( X = 1) P ( X 2) = 1 / 3 . Similarly, P ( X = 2 | X 2) = 2 / 3 . Thus, the conditional p.m.f. of X is Pr { X = k | X 2 } = 1 / 3 for k = 1 2 / 3 for k = 2 0 otherwise. Problem 3
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This note was uploaded on 02/01/2010 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Institute of Technology.

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hmwk5sol2027 - Solutions to Homework 5, ISyE 2027 Problem 1...

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